我使用wget，如果在mac或linux上，你必须在windows或终端中的cmd提示符中安装模块

pip install wget

这非常简单，只需使用download()函数即可

import wget
url = input("Enter Url to download: ")
wget.download(url)

TQDM也是一个选择，你也必须下载这个模块。

pip install tqdm

现在确保导入模块，设置范围和pass

from tqdm import tqdm
for i in tqdm(range(int(9e7))):
    pass

我喜欢这一页。

从简单的示例开始，然后转移到多线程版本。开箱即用。不需要第三方软件包。

代码看起来像这样:

import time
import sys

def do_task():
    time.sleep(1)

def example_1(n):
    for i in range(n):
        do_task()
        print '\b.',
        sys.stdout.flush()
    print ' Done!'

print 'Starting ',
example_1(10)

或者这里是一个使用线程的例子，以便在程序运行时运行旋转加载条:

import sys
import time
import threading

class progress_bar_loading(threading.Thread):

    def run(self):
            global stop
            global kill
            print 'Loading....  ',
            sys.stdout.flush()
            i = 0
            while stop != True:
                    if (i%4) == 0: 
                        sys.stdout.write('\b/')
                    elif (i%4) == 1: 
                        sys.stdout.write('\b-')
                    elif (i%4) == 2: 
                        sys.stdout.write('\b\\')
                    elif (i%4) == 3: 
                        sys.stdout.write('\b|')

                    sys.stdout.flush()
                    time.sleep(0.2)
                    i+=1

            if kill == True: 
                print '\b\b\b\b ABORT!',
            else: 
                print '\b\b done!',


kill = False      
stop = False
p = progress_bar_loading()
p.start()

try:
    #anything you want to run. 
    time.sleep(1)
    stop = True
except KeyboardInterrupt or EOFError:
         kill = True
         stop = True

我使用format()方法来制作一个加载条。以下是我的解决方案:

import time

loadbarwidth = 23

for i in range(1, loadbarwidth + 1):
    time.sleep(0.1) 

    strbarwidth = '[{}{}] - {}\r'.format(
        (i * '#'),
        ((loadbarwidth - i) * '-'),
        (('{:0.2f}'.format(((i) * (100/loadbarwidth))) + '%'))
    )

    print(strbarwidth ,end = '')

print()

输出:

[#######################] - 100.00%

我喜欢加布里埃尔的答案，但我改变了它的灵活性。您可以将bar-length发送到函数，并获得您想要的任何长度的进度条。进度条的长度不能为零或负。同样，你也可以像Gabriel answer一样使用这个函数(请看例子#2)。

import sys
import time

def ProgressBar(Total, Progress, BarLength=20, ProgressIcon="#", BarIcon="-"):
    try:
        # You can't have a progress bar with zero or negative length.
        if BarLength <1:
            BarLength = 20
        # Use status variable for going to the next line after progress completion.
        Status = ""
        # Calcuting progress between 0 and 1 for percentage.
        Progress = float(Progress) / float(Total)
        # Doing this conditions at final progressing.
        if Progress >= 1.:
            Progress = 1
            Status = "\r\n"    # Going to the next line
        # Calculating how many places should be filled
        Block = int(round(BarLength * Progress))
        # Show this
        Bar = "[{}] {:.0f}% {}".format(ProgressIcon * Block + BarIcon * (BarLength - Block), round(Progress * 100, 0), Status)
        return Bar
    except:
        return "ERROR"

def ShowBar(Bar):
    sys.stdout.write(Bar)
    sys.stdout.flush()

if __name__ == '__main__':
    print("This is a simple progress bar.\n")

    # Example #1:
    print('Example #1')
    Runs = 10
    for i in range(Runs + 1):
        progressBar = "\rProgress: " + ProgressBar(10, i, Runs)
        ShowBar(progressBar)
        time.sleep(1)

    # Example #2:
    print('\nExample #2')
    Runs = 10
    for i in range(Runs + 1):
        progressBar = "\rProgress: " + ProgressBar(10, i, 20, '|', '.')
        ShowBar(progressBar)
        time.sleep(1)

    print('\nDone.')

# Example #2:
Runs = 10
for i in range(Runs + 1):
    ProgressBar(10, i)
    time.sleep(1)

结果:

这是一个简单的进度条。 示例# 1 进度:[###-------]30% 例# 2 进步 : [||||||||||||........) 60% 完成了。

下面的代码是一个相当通用的解决方案，也有一个时间消耗和剩余时间估计。你可以使用任何可迭代对象。进度条的大小固定为25个字符，但可以使用完整、半块和四分之一块字符以1%的速度显示更新。输出如下所示:

 18% |████▌                    | \ [0:00:01, 0:00:06]

代码示例:

import sys, time
from numpy import linspace

def ProgressBar(iterObj):
  def SecToStr(sec):
    m, s = divmod(sec, 60)
    h, m = divmod(m, 60)
    return u'%d:%02d:%02d'%(h, m, s)
  L = len(iterObj)
  steps = {int(x):y for x,y in zip(linspace(0, L, min(100,L), endpoint=False),
                                   linspace(0, 100, min(100,L), endpoint=False))}
  qSteps = ['', u'\u258E', u'\u258C', u'\u258A'] # quarter and half block chars
  startT = time.time()
  timeStr = '   [0:00:00, -:--:--]'
  activity = [' -',' \\',' |',' /']
  for nn,item in enumerate(iterObj):
    if nn in steps:
      done = u'\u2588'*int(steps[nn]/4.0)+qSteps[int(steps[nn]%4)]
      todo = ' '*(25-len(done))
      barStr = u'%4d%% |%s%s|'%(steps[nn], done, todo)
    if nn>0:
      endT = time.time()
      timeStr = ' [%s, %s]'%(SecToStr(endT-startT),
                             SecToStr((endT-startT)*(L/float(nn)-1)))
    sys.stdout.write('\r'+barStr+activity[nn%4]+timeStr); sys.stdout.flush()
    yield item
  barStr = u'%4d%% |%s|'%(100, u'\u2588'*25)
  timeStr = '   [%s, 0:00:00]\n'%(SecToStr(time.time()-startT))
  sys.stdout.write('\r'+barStr+timeStr); sys.stdout.flush()

# Example
s = ''
for c in ProgressBar(list('Disassemble and reassemble this string')):
  time.sleep(0.2)
  s += c
print(s)

欢迎提出改进建议或其他意见。干杯!

这是我的简单解决方案:

import time

def progress(_cur, _max):
    p = round(100*_cur/_max)
    b = f"Progress: {p}% - ["+"."*int(p/5)+" "*(20-int(p/5))+"]"
    print(b, end="\r")

# USAGE:
for i in range(0,101):
    time.sleep(0.1) 
    progress(i,100)

print("..."*5, end="\r")
print("Done")